Which of the following compounds will be colored in the solid state?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(CuF_2\)
A transition metal compound is coloured when its metal ion has a partly filled d subshell. An electron can then be excited from a lower to a higher d level (a d–d transition); the energy absorbed lies in the visible region, and the compound shows the complementary colour. An ion with \( \mathrm{d^{0}} \) or a completely filled \( \mathrm{d^{10}} \) subshell cannot do this and is colourless.
In \( \ce{CuF2} \), x + 2(−1) = 0, so copper is +2. Cu (Z = 29) is \( \mathrm{[Ar]\,3d^{10}\,4s^{1}} \); losing the 4s electron and one 3d electron gives \( \ce{Cu^2+} \): \( \mathrm{3d^{9}} \), with one unpaired electron. So \( \ce{CuF2} \) is coloured in the solid state.
Why the other options are incorrect
Option (1) \(Ag_2SO_4\) — Incorrect. 2x + (−2) = 0, so silver is +1. Ag (Z = 47) is \( \mathrm{[Kr]\,4d^{10}\,5s^{1}} \); losing the 5s electron gives \( \ce{Ag+} \): \( \mathrm{4d^{10}} \), a full subshell, so the salt is white.
Option (3) \(ZnF_2\) — Incorrect. Zinc is +2. Zn (Z = 30) is \( \mathrm{[Ar]\,3d^{10}\,4s^{2}} \); losing the two 4s electrons gives \( \ce{Zn^2+} \): \( \mathrm{3d^{10}} \), so it is colourless.
Option (4) \(Cu_2Cl_2\) — Incorrect. 2x + 2(−1) = 0, so copper is +1. Losing only the 4s electron gives \( \ce{Cu+} \): \( \mathrm{3d^{10}} \), a full subshell, so it is white.